0,1

There are always at least two squares between positive consecutive cubes, starting with the perfect squares 1 and 4 between the perfect cubes 1 (included) and 8 (excluded).

The number of squares between the cube of n (included) and the cube of n+1 (excluded) is always one of the two integers bracketing 3*sqrt(n)/2.

The number a(n) in the sequence is 0 if the correct count is the lower number or 1 if the actual count is the higher number.

G. P. Michon, A Sequence of Bits with Strange Statistics.

a(n) = floor(sqrt((n+1)^3-1)) - ceiling(sqrt(n^3)) + 1 - floor(1.5 sqrt(n))

The sequence starts with a(0)=1 for n=0 because there is just one perfect square (0) between the cube of 0 (included) and the cube of 1 (excluded).

This exceeds by a(0)=1 the asymptotic expression floor(1.5*sqrt(n)) for the value n=0.

Adjacent sequences: A140071 A140072 A140073 this_sequence A140075 A140076 A140077

Sequence in context: A092079 A139312 A071041 this_sequence A090174 A165556 A127243

easy,nonn

Gerard P. Michon (g.michon(AT)att.net), May 06 2008